Answer:
<h3>The answer is 8.91 m/s²</h3>
Explanation:
The acceleration of an object given it's mass and the force acting on it can be found by using the formula
f is the force
m is the mass
From the question we have
We have the final answer as
<h3>8.91 m/s²</h3>
Hope this helps you
The force exerted on the board by the karate master given the data is -4500 N
<h3>Data obtained from the question </h3>
- Initial velocity (u) = 10 m/s
- Final velocity (v) = 1 m/s
- Time (t) = 0.002 s
- Mass (m) = 1 Kg
- Force (F) = ?
<h3>How to determine the force</h3>
The force exerted can be obtained as illustrated below:
F = m(v - u) / t
F = 1 (1 - 10) / 0.002
F = (1 × -9) / 0.002
F = -4500 N
Learn more about momentum:
brainly.com/question/250648
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Answer:
Explanation:
From the free-body diagram for the car, we have that the normal force has a vertical component and a horizontal component, and this component act as the centripetal force on the car:
Solving N from (2) and replacing in (1):
The centripetal acceleration is given by:
Replacing and solving for v:
Acceleration = (change in speed) / (time for the change)
Change in speed = (ending speed) - (starting speed)
Change in speed = (60 km/hr) - (50 km/hr) = 10 km/hr
Time for the change = 12 sec
Acceleration = (10 km/hr) / (12 sec)
Acceleration = 0.8333 km/hr-sec
Convert to a unit that we can understand:
Acceleration = (0.8333 km/hr-sec)x(1000 m/km)x(hr/3600sec)
Acceleration = (833.3 / 3600) (m-hr / hr-sec²)
Acceleration = (833.3 / 3600) (m/s²)
<em>Acceleration = 0.231 m/s² </em> or 0.833 km/hr-sec²
Answer:
The exit temperature of cold water is 37.6°C
Explanation:
Mass flow rate of water, 
Mass flow rate of air, 
Applying the conservation of heat law between air, water and the surrounding:
![[\dot{m}c(T_{in} - T_{out})]_{air} = [\dot{m}c(T_{in} - T_{out})]_{water} + \dot{Q}_{lost}](https://tex.z-dn.net/?f=%5B%5Cdot%7Bm%7Dc%28T_%7Bin%7D%20-%20T_%7Bout%7D%29%5D_%7Bair%7D%20%3D%20%5B%5Cdot%7Bm%7Dc%28T_%7Bin%7D%20-%20T_%7Bout%7D%29%5D_%7Bwater%7D%20%2B%20%5Cdot%7BQ%7D_%7Blost%7D)
We get the exit temperature of cold water by making 
the subject of the formula above:
![(T_{out})_{water} =\frac{ [\dot{m}c(T_{in} - T_{out})]_{air} + [\dot{m} c T_{in}]_{water} - \dot{Q}_{lost}}{(\dot{m}c)_{water}} \\\\(T_{out})_{water} = \frac{(2.5*1010*(90-20) + [1.2*4186*(8+273)] - 28000}{1.2*4186} \\\\(T_{out})_{water} = 310.6 K\\\\(T_{out})_{water} = 37.6^0 C](https://tex.z-dn.net/?f=%28T_%7Bout%7D%29_%7Bwater%7D%20%3D%5Cfrac%7B%20%5B%5Cdot%7Bm%7Dc%28T_%7Bin%7D%20-%20T_%7Bout%7D%29%5D_%7Bair%7D%20%2B%20%5B%5Cdot%7Bm%7D%20c%20T_%7Bin%7D%5D_%7Bwater%7D%20-%20%5Cdot%7BQ%7D_%7Blost%7D%7D%7B%28%5Cdot%7Bm%7Dc%29_%7Bwater%7D%7D%20%5C%5C%5C%5C%28T_%7Bout%7D%29_%7Bwater%7D%20%3D%20%5Cfrac%7B%282.5%2A1010%2A%2890-20%29%20%2B%20%5B1.2%2A4186%2A%288%2B273%29%5D%20-%2028000%7D%7B1.2%2A4186%7D%20%5C%5C%5C%5C%28T_%7Bout%7D%29_%7Bwater%7D%20%3D%20310.6%20K%5C%5C%5C%5C%28T_%7Bout%7D%29_%7Bwater%7D%20%3D%2037.6%5E0%20C)
